Solution:
Set $PT \cap BC = P_1$, $NT \cap AB = N_1$ and $MT \cap AC = M_1$. The triangles $N_1MT$, $PTM_1$ and $TP_1N$ are similar to $\triangle ABC$. Set $k_1 = \frac{N_1M}{AB}$, $k_2 = \frac{PT}{AB}$ and $k_3 = \frac{TP_1}{AB}$. Then
$$k_1 + k_2 + k_3 = 1$$
since
$$k_1 + k_2 + k_3 = \frac{N_1M}{AB} + \frac{PT}{AB} + \frac{TP_1}{AB} = \frac{N_1M}{AB} + \frac{AN_1}{AB} + \frac{MB}{AB} = 1$$

a) It is clear that $\frac{AM}{MB} = \frac{PT + N_1M}{AB} = \frac{k_1 + k_2}{k_3}$. Analogously, $\frac{BN}{NC} = \frac{k_1 + k_3}{k_2}$, $\frac{CP}{PA} = \frac{k_2 + k_3}{k_1}$. It follows by $\frac{AM}{MB} = \frac{BN}{NC}$ that $\frac{k_1 + k_2}{k_3} = \frac{k_1 + k_3}{k_2}$, i.e., $(k_2 - k_3)(k_1 + k_2 + k_3) = 0$. Hence $k_2 = k_3$. We get in the same way that $k_1 = k_2$ and then $k_1 = k_2 = k_3$. Hence $PT = TP_1$ and since $PP_1 \parallel AB$, it follows that the line $CT$ meets $AB$ at its midpoint. Analogously, the lines $BT$ and $AT$ meet $AC$ and $BC$ at their midpoints. Hence $T$ is the centroid of $\triangle ABC$.

b) We have
$$\begin{aligned} S_{MNP} & = S_{MNT} + S_{NPT} + S_{PMT} = S_{MBT} + S_{TNC} + S_{PAT} \\ & = \frac{1}{2}\left(S_{MBP_1T} + S_{TNC M_1} + S_{PAN_1T}\right) = \frac{S_{ABC}}{2}\left(1 - k_1^2 - k_2^2 - k_3^2\right) \end{aligned}$$
It follows by the inequality $k_1^2 + k_2^2 + k_3^2 \geq \frac{(k_1 + k_2 + k_3)^2}{3}$ and (1) that $k_1^2 + k_2^2 + k_3^2 \geq \frac{1}{3}$. Then
$$S_{MNP} \leq \frac{S_{ABC}}{2}\left(1 - \frac{1}{3}\right) = \frac{1}{3} S_{ABC}$$